Theory and applications of finite groups, by G.A. Miller, H. F. Blichfeldt [and] L. E. Dickson.

~ 73]

GROUPS OF ORDER paq1

185

factors of composition whenever there is a series of composition corresponding to every possible order of the factors of composition.
73. Groups of Order pa qf, p and q being Prime Numbers.
In ~ 138 it will be proved that every group of order p"qS is
solvable. Two very simple special cases will be considered
here. The case when = 1 is especially simple and will be
considered first. If G is of order paq and q <p, then G contains
only one subgroup of order pa and it must therefore be solvable.
When q>p and G contains more than one subgroup of order
p", it must contain just q such subgroups. We proceed to prove
that no two of these subgroups can have a cross-cut whose
order exceeds unity.
Let K represent the largest possible cross-cut of a pair of
these subgroups. Since K is invariant under operators of each
of these subgroups which are not contained in K, it must be
invariant under a subgroup of G whose order is divisible by
some prime number besides p. Hence K is invariant under an
operator of order q, and it is therefore contained in each one
of the q subgroups of order pa. Since K is composed of all the
operators which are common to a complete set of conjugates,
it is invariant under G, and the corresponding quotient group
has an order which is of the same form as the order of G.
It remains therefore only to consider the case when the q subgroups of order pa are such that no two of them have two
operators in common.
In this case these q subgroups are transformed according
to a transitive substitution group of degree q which involves no
substitution whose degree is less than q-1. It can therefore
not contain more than q-1 substitutions of degree q, since
the average number of letters in all its substitutions is q-1.
Hence it contains only one subgroup of order q, and it must
therefore be composite. As no group of order p"q can be
simple, every group whose order is of this form must be solvable.
The case when all the Sylow subgroups of a group of order
paqa are abelian is almost equally elementary. Let G be such
a group and suppose that G is simple. If s represents any